Problem: For a function $f$, we are given that $f(8)=1$ and $f'(8)=2$. What's the equation of the tangent line to the graph of $f$ at $x=8$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y-2=1(x-8)$ (Choice B) B $y-8=2(x-1)$ (Choice C) C $y-8=1(x-2)$ (Choice D) D $y-1=2(x-8)$
The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $f'(8)$ gives the slope of the tangent line to the graph of $f$ where $x=8$. We are given that $f'(8)=2$, so the slope of the tangent line is $2$. Furthermore, we are given that $f(8)=1$, which means the point of intersection of the tangent line and the graph is $(8,1)$. To summarize, the tangent line has a slope of $2$ and it passes through the point $(8,1)$. We can use the point-slope form of linear equations to find the tangent line equation: $\begin{aligned} y-y_1&=m(x-x_1) \\\\ y-1&=2(x-8) \\\\ \end{aligned}$ The equation is $y-1=2(x-8)$.